As is well known, Goldbach conjectured that every even positive integer (greater than 2) is the sum of two primes. While this is a difficult open problem, progress has been made from a number of different directions. Perhaps most notably, Chen has shown that every sufficiently large even positive integer is the sum of a prime and an almost prime (that is an integer that is a product of at most two primes). In another direction, Montgomery and Vaughan have shown that if is the set of positive even integers that cannot be expressed as the sum of two primes then

for some positive constant . This is stronger than the observation (which was made much earlier) that almost every positive integer can be expressed as the sum of two primes. In this post we’ll be interested in sets of integers with the property that most integers can be expressed as the sum of two elements from the set. To be more precise we’ll say that a set of positive integers has the Goldbach property (GP) if the sumset consists of a positive proportion of the integers. From the preceding discussion we have that the set of primes has the GP. (This discussion is closely related to the theory of thin bases.)

A natural first question in investigating such sets would be to ask how thin such a set can be. Simply considering the number of possible distinct sums, the reader can easily verify that a set (of positive integers) with the GP must satisfy

This is to say that a set of positive integers with the GP must satisfy for all large . Recall that the prime number theorem gives us that  for the set of primes . Thus the primes are much thicker than a set with the GP needs to be (at least from naive combinatorial considerations).

Considering this, one might ask if there is a subset of the primes with the GP but having significantly lower density in the integers. I recently (re)discovered that the answer to this question is yes. In particular we have that

 Theorem 1 There exists a subset of the primes such that has positive density in the integers and

 This last inequality shows that is as thin as such a set can be. In fact, one can show that a generic subset of the primes of this density will work. This result, in an more quantitative form, has been observed by Granville and earlier by Wirsing. Here I’ll present the somewhat different argument I found which is based on a theorem of Bourgain.

 Some properties of sets

In this section we define sets and record some of their properties. Let . For we say that is a set (with constant ) if for every function such that (here denotes the -th Fourier coefficient of ) we have

There is a rich theory to sets (being an instance of the Restriction phenomenon in Fourier analysis), however we will only need a few key properties. The following theorem was proved in this paper of Rudin (which is also where sets were first defined).

Theorem 2 If is a set for , and is an arithmetic progression then .  More specifically, we have that

This tells us, for example, that a set must satisfy for large . These sets also have a number of nice combinatorial properties. Here we record a basic lemma (which can be found as Lemma 4.30 in Tao and Vu’s Additive Combinatorics) which is easily deduced from Parseval’s identity.

Lemma 3 Let be a set with constant and let be a finite subset of . We then have

 We now state a theorem of Bourgain.

 Theorem 4 For any there exists a set, consisting of only of primes such that

 The proof of this theorem is fairly involved and relies on number theoretic as well as probabilistic arguments. We will only need this result in the case when . In this case (since is an even integer and norms can be expanded) both parts of the argument can be significantly simplified. In particular the only number theoretic information that is required are upper bounds on the number of ways an integer can be represented as the sum of two primes (see, for example, this paper of Green and Tao). Information of this sort can be readily obtained by standard sieve methods.

Let us now fit these pieces together to prove the theorem.

Proof of Theorem 1

Let be the set given by Bourgain’s theorem. By virtue of the fact that is a set we have that 

On the other hand, for sufficiently large we also have that . Applying Lemma 3, we have that . We can conclude that 

and the theorem follows.

Some Remarks

The argument here (via the proof of Bourgain’s theorem) constructs the set in a probabilistic manner. This is also the case in the arguments of Granville and Wirsing. In fact all of these arguments proceed by showing that a generic (in an appropriate sense) has the property that one is seeking. It would be very interesting to construct an explicit example of such a set.

As we remarked earlier we are only using Bourgain’s theorem when is an even integer, in which case his argument can be simplified considerably. It would be interesting to find an analogous combinatorial interpretation of sets for . Presumably this would lead to a deeper result about the primes.

The proof of Bourgain’s theorem uses very little about the primes. Similar ideas can be applied to other number theoretic sets, such as polynomial sequences. In fact, his paper gives a general criterion for a set to containing a subsets of maximal density.

Find the largest prime factor of .  Note that .

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La Ville de Nantes a exprimé début octobre, son souhait de sortir du conflit lié à la mise en place du passeport biométrique.
Ce n’est pas trop tôt après 6 mois de perturbations causées par l’absence de dialogue social !

Pour autant, au lieu de répondre aux préoccupations des agents, M. Bolzer Adjoint au Personnel de la Mairie de Nantes “demande que les agents renoncent au droit de grève dans ce service” (ndlr : il n’est pas demandé aux agents de terminer la grève, mais de “renoncer au droit de grève”, ce qui est une première en France !).
Et tout celà pour… 33,75 € par mois pendant 6 mois, soit 202 € pour solde de tout compte !
Ce n’est pas ce que souhaitent les agents qui revendiquent une prime pérenne correspondant à la technicité liée à la délivrance de ces passeports.

Lire l’article paru dans Presse-Océan du Mercredi 7 octobre 2009

Un décret du 17 avril 2008 institue une prime de restructuration de service applicable aux magistrats mais aussi aux fonctionnaires et agents non titulaires de l’Etat affectés dans des services faisant l’objet de restructurations.

Ces restructurations correspondent aux opérations de mutualisations entreprises depuis peu par la Ville de Nantes et par Nantes Métropole.

Un arrêté du 9 juillet 2008, applicable aux restructurations des tribunaux judiciaires, précise que les primes de restructuration de service sont accordées alors même que la distance entre l’ancienne et la nouvelle résidence administrative est inférieure ou égale à 20 km, ce qui est le cas entre la Ville de Nantes et Nantes Métropole.
Les montants de cette prime restent très hiérarchiques mais s’élèvent tout de même à :
- 9 000 € pour un Magistrat
- 5 000 € pour un Juge d’instance
- 2 000 € pour un Juge de proximité

Nous sollicitons donc des Elus de Nantes Métropole et de la Ville de Nantes, l’attribution de cette prime.

Voir le Décret du 17 avril 2008 et l’Arrêté du 9 juillet 2008

By showing that sums of this form are equivalent to products of this form (where p denotes the set of primes), Euler ablished a deep relation between the sum of the reciprocals of the powers of all positive integers and the set of prime numbers. Little did he know that the true profundity of this statement transcended all he ever thought was possible in evaluating patterns in the sequence of prime numbers. It was Euler himself who said that

“Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the mind will never penetrate.”

However in 1859, Bernhard Riemann took Euler’s finding, piled onto it the most sophisticated methods of mathematical analysis of his day and achieved a result which would change the way mathematicians approach prime numbers forever. In just a mere 8 pages of shitty handwriting, Riemann conveyed these ideas:

_

RIEMANN’S MAIN RESULT

By consequence of Euler’s statement, . Riemann was lazy. So he defined .
In mathematics, the logarithm is undefined at 0 for all bases. Riemann makes use of this property to bring out the zeroes of by taking the logarithm of it (by not defining them ironically), which also conveniently reduces the product of primes into a sum. Furthermore, Riemann notes the particularly neat form of which expands into a positive sum of reciprocals through Taylor expansion.
Summarily, expanding every term leads to .

Next, Riemann does a very cool thing in his paper by showing that such a sum may be expressed as an integral with respect to a certain function of x. The reason for this being that with integrals, powerful methods of analysis may be applied directly onto the problem at hand.
Imagine the sum where x runs over all positive integers instead of being isolated to the set of prime numbers. Such a sum would be equivalent to if the terms in which x was not a prime number were eliminated or equivalent to zero. Thus, such a sum would equal if when x is an n power of a prime p then would multiply that term by a factor of . If x is not a power of a prime, f(x) returns 0.
Observe:

Prior to this, Riemann also defined a prime counting function specifically catered for this paper he was writing. Define which increases by a value of 1 at every prime, at every prime square, at every prime cube et cetera. Thus , , , to name a few. Note that . This relation will play an important role in the integral Riemann will derive. Also, recall the definition of Gauss’ prime counting function, which increases by 1 over primes. The relation then becomes obvious after a few moments of concentrated thought. By the end of 8 pages of maths, Riemann’s main result was an explicit expression of R(x) in terms of the zeroes of . By providing an explicit formula for the value of R(x), it becomes clear that or the number of primes below a certain value can be explicitly given via such methods. To be precise can be made the subject of the equation via a method of inversion known as the the Möbius inversion. The inversion effectively takes the equation:

and turns it into

where the sign of each term depends on the prime factorization of the denominators of its coefficient.
For coefficients consisting of an odd number of prime factors the sign is negative.
For coefficients consisting of an even number of prime factors the sign is positive.
However for coefficients with repeated prime factors the sign is 0 and the term is omitted.

What Riemann did in his paper was to state that where R(x) is a sum over f(n) as long as . Essentially, so you can see it really is just direct substitution of f(x) for . Because R(x) is a step function, the area generated by the integral is exactly equivalent to the sum over discrete integers.

By defining in the form of this weird integral, Riemann uses integration by parts to express in terms of an integral of R(x) without the x axis having to increase in such an awkward manner. Believe me when I state because the reasoning behind the process is quite hard to explain although the general rule is simple. For all integrals of the form they are equivalent to as long as h(x) does not increase faster than x (that is to say on a graph with superimposed plots of both h(x) and x, the curve defined by h(x) will eventually be exceeded by the diagonal described by x).

Upon establishing , Riemann’s goal finally becomes clear. Through the use of the Fourier inversion theorem Riemann aims to express R(x) in terms of a contour integral of .

Given that

let , , and

And then it becomes clear that

Riemann evaluated the contour integral via overly complicated methods which I never managed to understand to produce this explicit formula:

where and are the non-trivial zeroes of the analytically continued

Subsequently, von Mangoldt simplified Riemann’s formula by expressing the Chebyshev function explicitly instead of Riemann’s function .
Originally defined by Russian mathematician Chebyshev in an attempt to simplify the Prime number theorem into something easier to handle, Chebyshev’s function is equivalent to where produces if n is a perfect power of prime p. Otherwise, . Note the similarity between and . This means that as , likewise . It’s importance shall be seen later. The statement that was also shown by Chebyshev himself to be equivalent to the Prime number theorem which was first proposed by Gauss. The Prime number theorem simply asserts that (see above) where the error term developed between and subsides to negligibility compared to the size of x. Riemann’s methods used to express prime counting functions in terms of the zeroes of the zeta function were ported over to the easier to handle which is asymptotically equal to x.

In many ways von Mangoldt’s explicit formula far exceeds Riemann’s original formula. For one thing there are no integrals in his explicit formula and for another it is much simpler to derive. The compensation is that all values for are transcendental whereas Riemann’s explicit formula for produces only rational numbers.

_

DERIVATION OF VON MANGOLDT’S VERSION OF THE EXPLICIT FORMULA
Von Mangoldt’s explicit formula was derived as such:

Differentiate to obtain
The key idea here is to see that because when the last bit of the integral resolves into something much nicer:
Then integrate by parts to obtain such that a Fourier inversion will produce

The next step then is to resolve into discrete terms which can be integrated one by one. Von Mangoldt starts off from the identity which can be easily derived from the reflection formula. The zeta resembling sign is in fact and the commonly misinterpreted term isn’t a product but . Note that the zeroes of are equivalent to the zeroes of thus the terms and are used interchangeably.

Take the logarithm of the identity and differentiate with respect to s to obtain:

By evaluating from the above formula and taking it away from the same formula, obtain

Therefore

Finally, by replacing with these four terms in the integral and evaluating it termwise, one obtains von Mangoldt’s result:

or

VISUALIZING VON MANGOLDT’S EQUATION

Von Mangoldt’s reformulation of Riemann’s main result in his seminal paper brings out the linearity of the function. As can be seen, the predominant x term essentially shows that the Prime number theorem is true provided the subsequent 3 terms tend to negligibility. The fourth term is quite obviously a non-changing constant and thus its involvement with the growth of is non-existent. The third term tends to zero because . The only possible term which might affect the growth of is the enigmatic second term which depends on the values of the complex zeores of .  In 1896 Hadamard showed that all non-trivial zeores (or all the complex zeroes) of can only have real parts within the range of 0 to 1 exclusive. The real component of the exponent of the zeroes being always smaller than 1 is sufficient to prove that the second term will tend to negligibility as x grows very large due to the slower growth of the second term (even though it’s smaller by an infinitesimal degree). In other words, if , then since the higher power in the denominator will eventually suppress the smaller power in the numerator. Conclusively, once the last three terms have been deemed to grow slower than x. This is essentially the outline of Hadamard’s proof of the Prime number theorem.

To show the literal equivalence between von Mangoldt’s derivation and the function , first estimate with only the real terms from von Mangoldt’s formula to give

Following that, add one zero term to the formula to give

where and denotes the first pair of non-trivial zeroes from the real axis (the complex zeroes occur in pairs known as complex conjugates which conveniently cancels the complexity of the other pair out to produce a real value)

Notice how the new term introduces a mild oscillation to the otherwise nearly linear line. This oscillation is actually predictable but when more zeroes get added the visible pattern breaks down into the chaos that is the pattern of the primes.

Adding 10 zero terms:

Notice how the estimate begins to coincide with at small x. The oscillation here is also predictable since only a finite number of zeroes have been added. But for every new zero term introduced the periodicity of the pattern extends over a larger range of x. Thus the periodicity will eventually extend out beyond the finite range of x shown in our graph giving the appearance of a lack of pattern. As the number of zero terms tend to infinity, the periodicity extends over an infinite range of x producing no recognizable pattern similar at all no matter how large we choose to zoom out our graph. At this point, the formula will have exactly traced out .

Here’s 50 zero terms and here’s 100. Notice how the extent of which is being sufficiently accurately approximated grows rightwards to cover more terms.

And just for sheer awesomeness, here’s 2000 zero terms added. The amplitudes of the oscillations at each horizontal bar gradually decrease while the frequency of the oscillations increase. At infinity, the amplitudes (or deviation from the actual value along the horizontal) become 0 while the infinitely rapid oscillations trace out every single point on the horizontal line. This is the literal equivalence of von Mangoldt’s derivation and the function . See here for more stuff.

Lors de cette rentrée, les enseignants titularisés en septembre 2009 à Paris vont pouvoir bénéficier de deux primes cumulables.

- la prime spéciale d’installation : versée à tous les professeurs des écoles titularisés dans certaines régions de France, dont l’île de France.

Les T1 de Paris bénéficient donc de cette prime d’un montant : 2039,31 €.

Les enseignants concernés vont recevoir un courrier avant fin septembre avec un formulaire à renvoyer au rectorat pour pouvoir bénéficier de la prime.

- la prime début de carrière : instaurée par le ministre de l’an dernier, cette prime de 1500 € sera versée en deux fois (750 euros seront versés en novembre 2009 et 750 euros en février 2010).

Elle sera versée automatiquement sur le salaire des enseignants concernés.

Textes de référence : décret n°2008-926 et arrêté du 12 septembre 2008

Après la pétition des cuisiniers des Etablissements d’Hébergement des Personnes Agées Dépendantes (EHPAD) du Centre Communal d’Action Sociale de Nantes, en faveur d’une prime pour les week-end de travail aux horaires décalés ou coupés, la CGT-FO a obtenu le 8 juillet 2009 une réponse de la Municipalité dans laquelle celle-ci s’est déclarée favorable à l’ouverture de négociations.

87 agents sont finalement concernés par ces horaires atypiques :
- 12 cuisiniers
- 40 auxiliaires de soins
- 35 agents sociaux

La revendication des cuisiniers, recueillie le 15 septembre par la CGT-FO, porte sur 21 € net par journée coupée.
Actuellement, la Municipalité nous propose une prime de 21 € brut par week-end.

Une pétition complémentaire est donc lancée auprès de l’ensemble des personnels concernés des EHPAD (contact : Sylvie au 02 40 41 95 47)

When a math lesson is over and there are still a few minutes available, I will sometimes ask a student for a number between 0 and 100.

This can spark a quick review of primes and the prime decomposition of a composite number.  Maybe the number is 25, 36 or another perfect square. Perhaps the number is 8 or 27 or another perfect cube.

My favorite is 42. As a science fiction fan, I know that 42 is the answer to “life, the universe, and everything,” according to Douglas Adams in his Hitchhikers Guide book series.

It just so happens that 42 is between the twin primes 41 and 43.  Numbers found between twin primes are multiples of six, with the exception of four — between 3 and 5, the first twin prime pair.

The pole of the Riemann Zeta function is a very fascinating thing and the following equation is true.

Or more correctly,

And so this is the case with primes:

And this is very weird: this.

Les promesses de la Municipalité nantaise faites avant les élections municipales se sont envolées :

La CGT-FO constate qu’aucune négociation sur les primes n’est ouverte à la Ville de Nantes et que seules 3 primes liées à des fonctions spécifiques ont été doublées (fossoyage, froid et extérieur).

La CGT-FO a rappelé à la Municipalité de Nantes, lors du CTP du 22 juin 2009, les revendications des personnels :
- alignement du régime indemnitaire “grade” de tous les agents sur celui de la Région des Pays de la Loire
- égalité des primes des personnels techniques entre Nantes Métropole et la Ville de Nantes
- création d’un régime indemnitaire “fonction” pour l’ensemble des autres filières (administrative, animation, culturelle, police municipale, sanitaire et sociale, sportive)

En outre, nous souhaitons que :
- les primes de fossoyage et de froid soient ajustées à 5,60 € comme pour les Agents de Surveillance de la Voie Publique (ASVP)
- les ASVP perçoivent une prime de risques de 16 % et se rapprochent ainsi des agents de la Police Municipale (PM)
- les primes liées à des fonctions spécifiques (fossoyage, froid, extérieur) soient forfaitisées et mensualisées

Les œuvres réalisées à Lyon sont parties au Brésil avec l’association “L’art de cœur de l’art” pour cette belle exposition internationale.

Merci Lucette!

Voir “Galerie (Brésil sept. 09)” Sur http://artdecoeurdelart

.com/

Collectif sur Pastel. Lyon 2008

Puis Institut Mackenzie jusqu’au 30 octobre 2009

PHILOSORAPTOR PHILOSOPHIZES

Here’s a simple and nice test for irreducibility in that N told me about a year ago. (I just noticed this lying around while cleaning; I don’t have a year’s buffer like Raymond Chen.) Apologies for the ugly formatting; you’ll have to trust that the result (Theorem 1, or Corollary is more beautiful than it looks.

Actually I’m not sure why I wrote this originally, given that it’s all already well-explained in the originals and even partially on Wikipedia. Perhaps my proofs are different or simpler or I was bored or something.

1. Irreducibility test

In its simplest form, the test can be stated as follows.

Theorem 1 Given a polynomial with integer coefficients, let . If there exists an integer such that is prime, then is irreducible.

It follows directly from the following two results:

Lemma 2 Given a polynomial with integer coefficients, let . Then for any root of , .

Proof: Let be any (complex) root of . As , either , or which means that . Thus in either case .

Lemma 3 Given a polynomial with integer coefficients, suppose is prime for some such that for every root of . Then is irreducible.

Proof: Suppose can be factorized in as where and have nonzero degree. As is prime, at least one of and must be 1, say .

Let the roots of (over ) be . Then , where the leading coefficient of being an integer, . As each factor , this implies that , which is a contradiction.

A related nice result can be proved:

Theorem 4 Let be a prime number, written in base as . Then the digit polynomial is irreducible in .

Note that this result does not automatically follow from Theorem 1. Unless , could be as large as , and the fact that is prime does not help, as the theorem only applies for .

On the other hand, Theorem 1 is “best-possible” in terms of the absolute values of the coefficients: consider, for , the polynomials , or , for both of which , and is prime!

Clearly, then, Theorem 4 depends not only on being prime, but also on the signs of the coefficients. It turns out that we only need the sign of :

Lemma 5 Given a polynomial with integer coefficients such that , let . Then for any root of , either or .

Proof: If , we are done. Else, as before, .

So if , then , so which means that , from which we get .

We can now prove the theorem for bases other than 2: Proof: As and all the coefficients lie between and , we have . Thus by Lemma 5, for any root of , we have if . As is prime, Lemma 3 now implies the irreducibility of .
The theorem is also true for base 2, see \cite{Murty} or \cite{BFO}.

2. Generalisations

We can consider a few variants of Lemma 2 that are stronger when .

Lemma 6 Given a polynomial with integer coefficients, let be the first nonzero coefficient of after (i.e., if ). Let . Then for any root of , .

Proof: If , we are done. Else, which means that . Thus in either case .

Lemma 7 Given a polynomial with integer coefficients, let . Then for any root of , .

Proof: If , we are done. Else, which means that , i.e., .

We can combine all these into the following result.

Corollary 8 Given a polynomial with integer coefficients and , let be the first nonzero coefficient of after (i.e., if ). Let . Let . If there exists an integer such that is prime, then is irreducible.

Proof: As in Lemmas 5, 6 and 7, we can prove that either or . Lemma 3 then gives the result.

Perhaps the “generalisations” in this section, as they contain no new idea, are like the “cheap generalization” Pólya talks about:

There are two kinds of generalizations. One is cheap and the other is valuable.

It is easy to generalize by diluting a little idea with a big terminology. It is much more difficult to prepare a refined and condensed extract from several good ingredients.

— [G. Pólya]

And also

However, one must not forget that there are two kinds of generalization, one facile and one valuable. One is generalization by dilution, the other is generalization by concentration. Dilution means boiling the meat in a large quantity of water into a thin soup; concentration means condensing a large amount of nutritive material into an essence. The unification of concepts which in the usual view appear to be far removed from each other is concentration. Thus, for example, group theory has concentrated ideas which formerly were found scattered in algebra, number theory, geometry and analysis and which appeared to be very different. Examples of generalizations by dilution would be still easier to quote, but this would be at the risk of offending sensibilities.

— [Pólya and Szegö]

3. Remarks and historical notes

4 was proved for base 10 by Arthur Cohn, a student of Issai Schur, as attributed by Pólya and Szegö in their book, first published in 1925.\cite{PolyaSzego}

It was extended to all bases by Brillhart, Filaseta and Odlyzko in 1981.\cite{BFO} They point out that not all irreducible polynomials can be shown irreducible by a theorem of this sort, as there exist irreducible polynomials that never take a prime value at integral arguments, e.g., .

A simplified proof was given by M. Ram Murty\cite{Murty}, who points out that this theorem is a (strong) converse to Buniakowski’s unproved conjecture, that any irreducible polynomial takes primes values infinitely often, unless it has a common divisor—this is a generalization of Dirichlet’s theorem about primes in arithmetic progressions. He also extends it to an analogue in function fields. He observes that this theorem sometimes applies when most traditional tests fail, e.g., is reducible modulo for every , but is prime which proves that it is irreducible.

4. References
[BFO] John Brillhart, Michael Filaseta, and Andrew Odlyzko. On an irreducibility theorem of A. Cohn. Canadian Journal of Mathematics, 33(5):1055–1059, 1981.
[Murty] M. Ram Murty. Prime numbers and irreducible polynomials. The American Mathematical Monthly, 109(5):452–458, 2002. [JSTOR]
[Polya] G. Polya. Induction and Analogy in Mathematics. Princeton University Press, New Jersey, 1954. Volume 1 of Mathematics and Plausible Reasoning.
[PolyaSzego] G. Pólya and G. Szegö. Problems and Theorems in Analysis. Springer–Verlag, Berlin, 1972.

CHL pointed out a gigantic flaw in the lemma of my failed proof of the prime number theorem.

This is what the lemma should be (or rather this was what Euler proved originally):

However it still remains true that . More on this later.

What is definitely true about the sum to prime n though is this:
Because
which I shall denote as equation 1 because I’m too lazy to write out the whole equation in the next section (also M denotes the Meissel-Mertens constant which is approximately 0.2614972128476427837554…)

Now don’t tell me that the brackets that should encompass all terms containing n are missing because I know that and I left them out intentionally. I really hate the sight of those stretched brackets. Anyway I don’t really see the point of those brackets here since every term on the left contains an n and thus your brain should be able to logically parse the meaning behind the notation.

The Meissel-Mertens constant is also defined as such (as a matter of fact this is the original definition):
When I saw this equation and equation 1 the first thought that crossed my mind was to increase the iteration of for equation 1. This is what I mean by increasing the iteration:
where and from equation 1. It turns out that the equality still holds no matter how high you increase the iteration of as long as the difference in iteration degree is 1. Rate of convergence similarly increases with iteration degree.

Thus

where x indicates iteration degree (define and ) and . It follows from the original definition of the Mertens’ constant that . Noting the asymptotic lower bounds of , and which are , and respectively, one sees that the degree of iteration is equivalent to the power of the logarithm.
Thus by extension, .

After seeing that a difference in one degree of iteration causes no change to the convergent value (M), naturally one wonders what happens when the difference in iteration degree becomes two or more. This brings me back to my original version of the lemma. The mistake I made was to state because I thought that where C is a constant. It turns out that C is indeed a constant (Mertens’ constant actually) due to a proof by CHL, although his proof uses a statement of the Prime number theorem. So in actuality my assumption was correct and my version of the lemma remains true. However, the truth of now cannot be used in a lemma to prove the Prime number theorem since it is essentially assuming the truth of the Prime number theorem to imply the truth of the Prime number theorem. Consequently my ‘proof’ (of the PNT) doesn’t prove anything.

Before CHL produced the proof to I had doubts about the convergence of . After he informed me of the gaping hole in my lemma I realised that the difference may not actually converge since this statement says that one prime term is added for every single integer increase in n. Compare this with which adds lesser and lesser prime terms per arbitrary increment in n as n progresses towards infinity. Although it is known that the latter expression equates to Mertens’ constant, the much more rapid sum of prime terms in the prior statement lead me to reconsider the convergence of the limit.

Given the definitions of Euler’s constant and Mertens’ constant , form the equation:

Now define and .

Therefore

However

where x may or may not converge (by the way also define x(r) to mean the value of the sum to r, and thus x(∞) = x)
Consequently subtracting the second equation from the first equation produces . However since is by definition, then . Originally I thought that the plot of x(r) versus r was a logarithmic function which increased without bound and thus caused the difference to diverge, however CHL’s proof implies that x(r) tends to an asymptotic limit as . Because of the proof , x is consequently M in this case. Furthermore, see plot.

Also from the equation formed from the definitions of Euler’s constant and Mertens’ constant, one eventually derives Mertens’ third theorem.

P.S. I give up proving the Prime number theorem entirely after being informed of Euler’s similar (and correct) conclusion from the result he obtained in the lemma. What I don’t understand is why Euler never published those findings on prime numbers and why Gauss or any other mathematician never saw that as potential proof. On a slightly unrelated note, CHL’s idea for proving the Prime number theorem is interesting and I shall henceforth dedicate effort into analyzing his insights.

I have at least four planned posts left in the series on symmetric functions, but unfortunately I’ll be very busy for the next two weeks. In the meantime, here are some thoughts on the primes.

Euclid’s proof of the infinitude of the primes is often held up as a shining example of mathematical proof. (Whether this reputation is deserved is a matter of opinion.) Euler’s proof via the zeta function is also classic. An even showier proof by Furstenburg is phrased in the language of topology. Today I’d like to share my personal favorite proof and discuss one of its possible consequences.

Special cases of Dirichlet’s theorem

Euclid’s proof generalizes to certain special cases of Dirichlet’s theorem. The general recipe is as follows. First, one finds an integer polynomial with the property that its integer values, with finitely many exceptions, are only divisible by primes in some congruence class. For example, the polynomial is only divisible by primes congruent to . Then one uses the following result.

Theorem: The set of primes dividing the integer values of an integer polynomial is infinite.

The standard proof of this result is a generalization of Euclid’s proof: if , one supposes that there are finitely many such primes and considers . After dividing out by , we obtain a number coprime to ; contradiction. One obtains Euclid’s result with and Dirichlet’s theorem for primes congruent to using the cyclotomic polynomials ; I’ve written about this proof in an old post. A paper of Conrad discusses the generalization of this method to primes congruent to where ; it is known that one can do no better by this method.

Polynomials are not important

The limitation of this method is that is required to be a polynomial, and the reason I no longer like the above proof is that the highly “algebraic” method does not suggest the following generalization at all.

Theorem: Let be a sequence of positive integers which grows slower than for every (in particular, any sequence of polynomial growth rate). Then the set of primes dividing some is infinite.

Proof. We prove the following more precise converse: any sequence of positive integers divisible by at most distinct primes grows at least as fast as .

But this is almost obvious. Let denote the number of numbers of the form less than or equal to . Clearly any such number must satisfy

and it follows that . In particular, for every . (This is not the strongest bound, but it’s enough.) It follows that

.

This is equivalent to the claim we wanted to prove. (Note that if is required to be strictly increasing we get an honest inequality ; otherwise, we can still say .)

I find this proof the best motivated out of any of the proofs I know, especially because ultimately it reduces to simple counting. It is also constructive in the sense that a failure to satisfy the above inequality for some (for a strictly increasing sequence) implies that you have already hit at least prime factors.

Motivated by the above result, a positive solution to the following problem would imply a proof of Dirichlet’s theorem that totally avoids the machinery of zeta functions.

Problem: Explicitly describe a sequence growing slower than for every that is divisible, with finitely many exceptions, only by primes in a specified arithmetic progression.

Of course we already know by the strong form of Dirichlet’s theorem that this is true of the sequence consisting precisely of the numbers divisible by primes in a specified arithmetic progression. But this is a “high-complexity” sequence whereas, say, the integer values of a cyclotomic polynomial are “low-complexity.” In the spirit of Polymath4 what would be nice is a sequence describable in polynomial time, but as long as the proof that both of the above conditions are met is elementary it really doesn’t matter. Ideally would be some perturbation of a polynomial, so the first condition would be trivial to verify. I have no idea how one would go about constructing such a sequence.

de Pierre Bourdieu

Un extrait du texte paru dans le monde diplomatiqueintitulé: “l’essence du néolibéralisme” , un peu ardu mais qui analyse bien les proccessus d’asservissement et de tromperies organisé du Système capitaliste libérale.

L’extrait:

…/…

Ainsi s’instaurent le règne absolu de la flexibilité, avec les recrutements sous contrats à durée déterminée ou les intérims et les « plans sociaux » à répétition, et, au sein même de l’entreprise, la concurrence entre filiales autonomes, entre équipes contraintes à la polyvalence et, enfin, entre individus, à travers l’ individualisation de la relation salariale : fixation d’objectifs individuels ; entretiens individuels d’évaluation ; évaluation permanente ; hausses individualisées des salaires ou octroi de primes en fonction de la compétence et du mérite individuels ; carrières individualisées ; stratégies de « responsabilisation » tendant à assurer l’auto-exploitation de certains cadres qui, simples salariés sous forte dépendance hiérarchique, sont en même temps tenus pour responsables de leurs ventes, de leurs produits, de leur succursale, de leur magasin, etc., à la façon d’« indépendants » ; exigence de l’« autocontrôle » qui étend l’« implication » des salariés, selon les techniques du « management participatif », bien au-delà des emplois de cadres.

Autant de techniques d’assujettissement rationnel qui, tout en imposant le surinvestissement dans le travail, et pas seulement dans les postes de responsabilité, et le travail dans l’urgence, concourent à affaiblir ou à abolir les repères et les solidarités collectives (3).

Pour Allez plus loin…

Déposséder les possédants.

La grève générale aux « temps

héroïques » du syndicalisme

révolutionnaire (1895-1906)

Textes rassemblés et présentés par Miguel Chueca

Le réseau anglais de Radio-Canada, CBC, a appris que 72 cadres de Nortel se sont partagé des augmentations en salaires et en bonus de 7,5 millions de dollars américains en 2009. De ce nombre, 14 vont recevoir des augmentations de plus de 0,5 M$ US.

1,68 million pour le dirigeant

Le plus important bénéficiaire de ce nouveau plan de compensation est l’ancien trésorier John Dolittle, qui a pris les rênes de l’entreprise en août dernier, après le départ de l’ancien directeur Mike Zafirovski. M. Dolittle voit ses compensations s’élever à 1,68 million, soit une augmentation de 1,12 million par rapport à 2008. Ce nouveau plan de compensation s’ajoute aux 45 millions de dollars versés à titre de prime de rétention et de bonus depuis le printemps dernier.

En juin dernier, Mike Zafirovski avait défendu sa décision de verser ces primes devant le comité des finances de la Chambre des communes, tout en refusant de verser des indemnités de départ à des centaines d'employés mis à pied et en diminuant les pensions versées aux retraités. M. Zafirovski a aussi tenté de faire valoir sa qualité de créancier devant une cour américaine en octobre dernier, réclamant 12 millions à l’entreprise, dont la moitié à titre de prestations de retraite. Sa demande est en examen.

Des créanciers non protégés

L’entreprise canadienne a déclaré faillite en janvier dernier et, de ce fait, a refusé de payer une indemnité de départ à plusieurs employés licenciés. Elle compte aussi mettre fin au versement d’indemnités à 410 employés en invalidité à long terme. Enfin, elle souhaite continuer de réduire les retraites de quelque 70 000 anciens employés.

Le déficit de l’entreprise était évalué entre 2500 et 2800  M$ en janvier dernier.

Nortel : Une faillite payante pour les cadres | Radio-Canada.ca

Il y a les primes au rendement et il y a les primes pour conserver les employés qu’on pourrait aussi appeler primes à la faillite pour les cadres supérieurs uniquement. Non seulement ils touchent quand ca marche bien mais ils touchent aussi des primes quand l’entreprise coule. Incroyable mais vraie !

I wanted to write this post yesterday, but I couldn’t work out some of the steps that I wanted to be able to work out, so I had to wait a day and try again. But I think I’ve got it now, let’s see…

We’ve been working with the function

Last time, we arrived (skipping over some rather large gaps) at an analytic formula for , namely:

where (mod issues)

The goal today is to use this formula for to find a formula for primes. In particular, we will find a formula for , which is the number of primes no larger than . These two functions are related, as follows: recall that is a step function which jumps by at every prime power . Pick an . How many jumps of size 1 does make before getting to ? Well, it jumps by 1 at the primes, so it makes jumps of size 1. How many jumps of size 1/2? These are jumps at prime squares, . This inequality is the same as , so we make jumps of size 1/2. Similarly, we’ll make jumps of size to calculate . Eventually (for some large enough ) will be less than 2, and so , and we can stop looking for jumps.

Translating the above into symbols, we have just decided that

I’ve written it as an infinite sum, but eventually (like I said) those values will be 0, so we’ve really got a finite sum.

Now wait, our goal was a formula for , in terms of , and I’ve got a formula the other way, in terms of .

This is the part I got stuck on yesterday. Edwards’ book says (or perhaps doesn’t, causing my problems) that basically you do Möbius inversion to the sum above, and obtain a formula for in terms of . He says slightly more than that, but I wasn’t able to piece it together as easily as he seemed to indicate it should work. But anyway, what’s this Möbius inversion thing? I’ll take a pretty low-level account, leaving lots of room for things to be jazzed up (which I may return to and do sometime soonish).

An integer is said to be “squarefree” if it factors into the product of distinct primes or, equivalently, it is not divisible by a square. Define the following function, called the Möbius function, on positive integers ( is supposed to indicate a prime):

So, for example, is -1 at any prime, , and .

Möbius inversion is then the following:

If then , and conversely.

Well, I tried to figure out how my expression for , in terms of s was such an of . I don’t see it, as stated. Perhaps somebody can point me in the right direction in the comments.

The observation you are supposed to make comes when you consider something like . Using our formula above, we can write

or, more generally, we have

If we subtract this equation (with ) from the equation for in terms of s, we can eliminate one term from the right-hand side. The idea is to iterate this process. I think a (reasonably, by my blog’s standards) clean way to write what is happening is as follows (it helps when you know the answer you are supposed to get ):

Consider the sum

which by our most recent formula is the same as

Now, fix a positive integer . The coefficient of in this last double sum is

Almost there… what is ? Turns out, it is 1 if (which is clear), and 0 otherwise. Indeed, suppose has prime factorization . Then all of the divisors, in the sum we’re after, are of the form where each . The only divisors, , where are those where the are all either 0 or 1. So the worthwhile could be 1, or some , or a (), or (, , ), etc. So

(By the way, this is (almost verbatim) the proof in Hardy and Wright’s “An Introduction to the Theory of Numbers”, and I’ll count that as one of my references for anything I say about ).

So, let’s gather back up here. We’ve now shown that in the sum

the coefficient of is 0 unless , when the coefficient is 1. Therefore, we have obtained out formula for :

I guess the point of Riemann’s paper was that the terrible integral formula he obtained for could then be used in this expression for , to give an exact analytic expression (that is, one only an analyst would find pretty) for . Pretty wild, in my mind.

It seems that if you get some feeling for the terms in , you can decide that the “dominant” term is the term. And then that still ends up as the dominant term in the expression above for . And so you get to say

So you’ve got that going for you, which is nice.